A particle of mass 0.350 kg is attached to the 100-cm mark of a meterstick of ma

Question

A particle of mass 0.350 kg is attached to the 100-cm mark of a meterstick of mass 0.150 kg. The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 2.00 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50.0-cm mark. 0.325 The moment of inertia of several objects is the sum of the moments of inertia of the individual objects, kg m^2/s Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark. kg m^2/s

Explanation / Answer

let

M = 0.35 kg

L = 1 m

m = 0.15 kg

a) angular momentum of the system = I*w

= ( M*L^2/12 + m*(L/2)^2 )*w

= ( 0.35*1^2/12 + 0.15*(1/2)^2 )*2

= 0.133 kg.m^2/s

b) angular momentum of the system = I*w

= ( M*L^2/3 + m*L^2 )*w

= ( 0.35*1^2/3 + 0.15*1^2 )*2

= 0.533 kg.m^2/s

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